Question

An article reported that for a sample of 54 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 162. a) Calculate and interpret a 95% two sided answers to two decimal places.) confidence interval for true average CO2 level n the population of a homes from hich the sample as selected Round your ppm Interpret the resulting interval o we are 95% confident that the true population mean lies above this interval. we are 95% confident that this interval does not contain the true population mean. o we are 95% confident that the true population mean lies below this interval O we are 95% confident that this interval contains the true population mean. (b) Suppose the investigators had made a rough guess of 181 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 60 ppmm for a confidence level of 95967 (Round your answer up to the nearest whole number) kitchens You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

a.)

For 95% confidence level, t = 2.005

n = 54

s = 162

$\bar{x}$ = 654.16

Lower - Upper bound: X+1x . l / bound:

Using above formula

Lower bound, L = 654.16 - 2.005 * ( 162 / 7.348)

L = 654.16 - 44.203

L = 609.957

Upper bound, U = 654.16 + 2.005 * ( 162 / 7.348)

U = 654.16 + 44.203

U = 698.363

Option D is correct. We are 95% confident that True population mean lies in this interval.

b.)

Given, s = 181

Width or Margin of Error, MOE = 60

t = 2.005

MOE = t * s / $\sqrt{n}$

60 = 2.005 * 181 / $\sqrt{n}$

n = 36.583 = 37

n = 37

Note: I used t-value to solve the problem. In case if you use z-statistic, z = 1.96 then your answer might differ